Lesson 33 - Kinetic theory and Pressure
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ALL
1. (a)
What do you see when looking at Brownian motion in air?
……… smoke particles that appear as tiny points of light in random motion (1) due to
collisions with air molecules (1) ………………………………………………………… (2)
(b)
What changes do you see if the air is heated?
…… the random motion becomes more violent ……………………………………… (1)
2. Define pressure and give its units.
………………………………………………………………………………………………...
P (Pa) = F (N) / A (m2)
………………………………………………………………………………………………...
…………………………………………………………………………………………… (2)
3. List six assumptions that you have to make about molecules when deriving the
equation for the kinetic theory of gases.
1
A Gas consists of particles called molecules.
2
The molecules are in constant random motion. As many travelling in one direction
as any other. The centre of mass of the gas is at rest.
3
Intermolecular forces are negligible.
4
The duration of collisions between molecules is negligible.
5
Molecules move with constant velocity in between collisions.
6
The volume of gas molecules is negligible compared with the volume of the gas.
7
All collisions are totally elastic.
8
Newtonian mechanics can be applied to the collisions.
……………………………………………………………6 of the above………… (6)
MOST
4.
(a)
a molecule of mass 3x10-26 kg moving at 4x102 ms-1 collides
elastically with a wall. What is the change of momentum of the molecule?
mv
= 3x10-26(4x102 – (- 4x102))
= 3x10-26x8x102
= 2.4x10-23 Ns
Change in momentum = ……2.4x10-23 ……………. Unit… Ns ……… (3)
How long would it take this molecule to travel a distance of 0.4 m?
t
(b)
= s/v
= 0.4/400
Time = ……0.01……….. s (2)
5.
(a)
What is meant by the root mean square speed of gas molecules?
……… the square root of the sum of the squares of the speeds of the molecules divided by the
number of molecules in the gas …………………………………...
………………………………………………………………………………………………...
………………………………………………………………………………………………...
…………………………………………………………………………………………… (2)
(b)
what is the difference between the root mean square of a group of
gas molecules and their average speed?
Example
Consider four molecules with the following speeds :
200 ms-1 300 ms-1 400 ms-1 500 ms-1
R.M.S = [(2002+3002+4002+5002)/4]1/2 = 367 ms-1
Mean speed = [200 + 300 + 400 + 500]/4 = 350 ms-1
The r.m.s. speed is greater than the mean speed.…………………………………… (2)
6.
Calculate the root mean square speeds for the following gases at a pressure
of 10 Pa: USE P = 1/3 v2 to give: v2 = 3P/
(a)
air
density 1.29 kgm-3
5
(b)
(c)
(d)
(e)
carbon dioxide
root mean square speed = …482………….. ms-1 (2)
density 1.98 kgm-3
nitrogen
root mean square speed = …398………….. ms-1 (2)
density 1.25 kgm-3
chlorine
root mean square speed = ……490 ……….. ms-1 (2)
density 3.21 kgm-3
hydrogen
root mean square speed = …306 ………….. ms-1 (2)
density 0.09 kgm-3
root mean square speed = ……1826 ……….. ms-1 (2)
5. Calculate the pressure of three samples of air of density 1.29 kgm-3 with different
root mean square speeds:
USE P = 1/3 v2
(a) 4.0x102 ms-1
pressure = ……6.88x104 ………..Pa (2)
(b) 5.0x102 ms-1
pressure = …1.075x105 …………..Pa (2)
2
(c) 6.0x10 ms
-1
pressure = …1.548x105 …………..Pa (2)
9.
A sample of gas of volume 0.1 m3 and at a pressure of 2.0x105 Pa is
enclosed in a cylinder. If the root mean square speed of the gas molecules is 5.5x102
ms-1 and the mass of each molecule is 3.5x10-26 kg calculate the number of gas
molecules in the cylinder.
PV = 1/3 mNv2
So
N
= 3PV/mv2
= 3x2x105x0.1/[3.5x10-26x3.025x105]
And
N
= 6x104/1.059x10-20
= 5.67x1024
Number of gas molecules = ……5.67x1024………… (3)
10.
A sample of gas at a pressure of 1.5x105 Pa is contained in a cylinder with a
volume of 0.05 m3. If there are 1.2x1025 molecules of gas in the cylinder with a root
mean square speed of 350 ms-1 calculate the mass of one gas molecule.
PV = 1/3 mNv2
So,
m
And
m
= 3PV/Nv2
= 3x1.5x105x0.05/[1.2x1025x1.225x105]
= 2.25x104/1.47x1030
= 1.53x10-26 kg
Mass of one gas molecule = ……1.53x10-26
…………kg (3)
SOME
11.
Suppose there are N molecules in a
rectangular box of dimensions a, b and l and
suppose that the molecule has a velocity v with
the components as shown.
vy
b
a)
What is the change in velocity
of the particle when it hits the shaded face if
the collision is totally elastic?
vx
vz
2mvx
a
(1)
b)
i)
What is the time interval
before the same molecule makes a 2nd collision
at the same face?
2Lvx.
L
(1)
ii)
An therefore what is the frequency of collisions?
vx/2L
(1)
c)
From the definition of Pressure and by finding the rate of change of
momentum for one molecule show that for N molecules:
P  Nmv2 / 3V .
Rate of change of momentum at the face = 2mvx x vx/2L= mvx2/L
(Since F=Δp/Δt)
(1)
Therefore the pressure exerted on the shaded face is by one molecule is:
P=F/A
(1)
P = (mvx2/L / (ab)
(1)
(ab = A, Area of shaded face)
If we sum N contributions, one from each particle in the box, each contribution
proportional to vx2 for that particle, the sum just gives us N times the average value
of vx2. That is to say,
P  F / A  Nmvx2 / LA  Nmvx2 / V
(1)
where there are N particles in a box of volume V. Next we note that the particles
are equally likely to be moving in any of 3 directions, so the average value of vx2
must be the same as that of vy2 or vz2, and since v2 = vx2 + vy2 + vz2, it follows that
(1)
P  Nmv2 / 3V .
(1)
(7)